Question: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $21.8$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living longer than $25.1$ years.
Solution: $21.8$ $18.5$ $25.1$ $15.2$ $28.4$ $11.9$ $31.7$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $21.8$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $18.5$ years and one standard deviation above the mean is $25.1$ years. Two standard deviations below the mean is $15.2$ years and two standard deviations above the mean is $28.4$ years. Three standard deviations below the mean is $11.9$ years and three standard deviations above the mean is $31.7$ years. We are interested in the probability of a porcupine living longer than $25.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $18.5$ years and the other half $({16\%})$ will live longer than $25.1$ years. The probability of a particular porcupine living longer than $25.1$ years is ${16\%}$.